Question 1209652: GetThere Airlines currently charges $200$ dollars per ticket and sells $40{,}000$ tickets a week. For every $10$ dollars they increase the ticket price, they sell $300$ fewer tickets a week. How many dollars should they charge to maximize their total revenue?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to determine the ticket price that maximizes revenue for GetThere Airlines:
1. **Define variables:**
* Let 'x' be the number of $10 price increases.
* Let 'P' be the ticket price.
* Let 'Q' be the number of tickets sold.
* Let 'R' be the total revenue.
2. **Express price and quantity in terms of x:**
* Price: P = 200 + 10x
* Quantity: Q = 40000 - 300x
3. **Write the revenue function:**
Revenue is price times quantity:
R(x) = P * Q
R(x) = (200 + 10x)(40000 - 300x)
4. **Expand the revenue function:**
R(x) = 8000000 - 60000x + 400000x - 3000x²
R(x) = -3000x² + 340000x + 8000000
5. **Find the vertex of the parabola:**
Since the revenue function is a quadratic with a negative leading coefficient, it represents a parabola opening downwards. The vertex will give the maximum revenue. The x-coordinate of the vertex is given by:
x = -b / 2a
x = -340000 / (2 * -3000)
x = -340000 / -6000
x = 56.67 (approximately)
6. **Determine the optimal price:**
Substitute the value of x back into the price equation:
P = 200 + 10x
P = 200 + 10 * 56.67
P = 200 + 566.7
P = $766.70
7. **Round to the nearest dollar:**
Since we're dealing with money, we should round the price to the nearest dollar. In this case, since x represents the number of $10 increases, we should use x = 56 or x = 57.
If x = 56, P = 200 + 10*56 = $760.
If x = 57, P = 200 + 10*57 = $770.
Let's calculate the revenue for both:
R(56) = (760)(40000 - 300*56) = 760 * 23200 = $17,632,000
R(57) = (770)(40000 - 300*57) = 770 * 22900 = $17,633,000
So, the maximum revenue occurs at $770.
**Answer:** They should charge $770 per ticket to maximize their total revenue.
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